Respuesta :
Answer:
0.4
Step-by-step explanation:
First Machine(M1) Produces 40% of the product.
P(M1)=40%=0.4
The first Machine Produces defective product(D) 2% of the Time.
P(D)=2%=0.02
Given a defective product(D), we want to find the probability that it was produced on the first machine(M1).
P(M1|D)=P(M1∩D)/P(D)
=(0.4 X 0.02)/0.02
=0.4
Answer:
Given a defective product, the probability it was produced on the first machine is [tex] \\ P(F|D) = 0.25[/tex].
Step-by-step explanation:
This is a case that can be solved using the Bayes' Theorem, which implies conditional probabilities.
We have to remember that in conditional probabilities, the sample space is replaced by the probability of a given event (as can be seen below).
First, we need to find the probability of producing a defective product:
[tex] \\ P(D) = P(D|F)*P(F) + P(D|S)*P(S)[/tex] (1)
Where
P(D), the probability of producing a defective product.
P(F), the probability that the product was produced by the first machine. In this case, P(F) = 0.40.
P(S), the probability that the product was produced by the second machine. In this case, P(S) = 0.60.
P(D|F), a conditional probability that the product is defective given (or assuming) that it was produced by the first machine. In this case, from the question, P(D|F) = 0.02.
P(D|S), a conditional probability that the product is defective given (or assuming) that it was produced by the second machine. In this case, from the question, P(D|S) = 0.04.
So, from formula (1):
[tex] \\ P(D) = P(D|F)*P(F) + P(D|S)*P(S)[/tex]
[tex] \\ P(D) = 0.02*0.40 + 0.04*0.60[/tex]
[tex] \\ P(D) = 0.032[/tex]
That is, the probability of producing a defective product from these two machines is P(D) = 0.032.
From conditional probabilities we know that:
[tex] \\ P(A \cap B) = P(A|B)*P(B)[/tex]
[tex] \\ P(A \cap B) = P(B|A)*P(A) = P(B \cap A) [/tex]
So
[tex] \\ P(A|B)*P(B) = P(B|A)*P(A)[/tex]
In the same way, we can say that
[tex] \\ P(F|D)*P(D) = P(D|F)*P(F) = P(D \cap F) = P(F \cap D)[/tex]
In words, the probability that both events (D and F) occur is given by the "relationship" of conditional probabilities between these two events.
Then
The probability that a product comes from the first machine, given it is  defective is as follows:
[tex] \\ P(F|D)*P(D) = P(D|F)*P(F)[/tex]
[tex] \\ P(F|D) = \frac{P(D|F)*P(F)}{P(D)}[/tex]
[tex] \\ P(F|D) = \frac{0.02*0.40}{0.032}[/tex]
[tex] \\ P(F|D) = 0.25[/tex] or 25%.
